Kirchhoff's Mesh Law or Volta2e Law (KVU

It srares as follows :
the algebraic sum of the products of currents and resistances in each of the conductors in any
closed path (or mesh) in a network plus the algebraic sum of the e.mjs. in that path is zero.
In other words, 1:IR + 1:e.mj. = 0 ...round a mesh
It should be noted that algebraic sum is the sum which takes into account the polarities of the
voltage drops.
The basis of this law is this : If we.start from a particular junction and go round the mesh till we
come back to the starting point, then we must be at the same potential with which we started. Hence,
it means that all the sources of e.m.f. met on the way must necessarily be equal to the voltage drops
in the resistances, every voltage being given its proper sign, plus or minus.
2.3. Determination of Voltage Sign
In applying Kirchhoff's laws to specific problems, particular attention should be paid to the
algebraic signs of voltage drops and e.m.fs., otherwise results will come out to be wrong. Following
sign conventions is suggested :
(a) Sign of Battery E.M.F.
A rise in voltage should be given a + ve sign and afall in voltage a -ve sign. Keeping this in
mind, it is clear that as we go from the -ve terminal of a battery to its +ve terminal (Fig. 2.3), there
is a rise in potential, hence this voltage should be given a + ve sign. If, on the other hand, we go from
+ve tenninal to -ve tenninal, then there is afall in potential, hence this voltage should be preceded by a -ve sign. It is important to note that the sign of the battery e.m.! is independent of the direction
of the current through that branch.
(b) Sign of IR Drop
Now, take the case of a resistor (Fig. 2.4). If we go through a resistor in the same direction as the
current, then ther is a fall in potential because current flows from a higher to a lower potential..
Hence, this voltage fall should be taken -ve. However, if we go in a direction opposite to that of the
current, then there is a rise in voltage. Hence, this voltage rise should be given a positive sign.
It is clear that the sign of voltage drop across a resistor depends on the direction of current
through that resistor but is independent of the polarity of any other source of e.m.! in the circuit
under consideration. Assumed Direction of Current Fig. 2.5
In applying Kirchhoff's laws to electrical networks, the question of assuming proper direction
of current usually arises. The direction of current flow may be assumed either clockwise or
anticlockwise. If the assumed direction of current is not the actual direction, then on solving the
quesiton, this current will be found to have a minus sign. If the answer is positive, then assumed
direction is the same as actual direction (Example 2.10). However, the important point is that once
a particular direction has been assumed. the same should be used throughout the solution of the
question.
Note. It should be noted that Kirchhoff's laws are applicable both to d.c. and a.c. voltages and
currents. However, in the case of alternating currents and voltages, any e.m.f. of self-inductance or
that existing across a capacitor should be also taken into account.Solving Simultaneous Equations
Electric circuit analysis with the help of Kirchhoff's laws usually involves solution of two or
three simultaneous equations. These equations can be solved by a systematic elimination of the
varia~les but the procedure is often lengthy and laborious and hence more liable to error. Determinants
and Cramer's rule provide a simple and straight method for solving network equations through
manipulation of their coefficients. Of course, if the number of simultaneous equaitons happens to be
very large, use of a digital computer can make the task easy.
2.6. Determinants
The symbol I ~ ~ I is called a determinant of the second order (or 2 x 2 ~eterminant) because
it contains two rows (ab and cd) and two columns (ac and bd). The numbers a, b, c and d are called
the elements or constituents of the determinant. Their number in the present case is 22=4.
The evaluation of such a determinant is accomplished by cross-multiplicaiton is illustrated